What does a superposition of states mean in quantum mechanics?

[mike1000]

[Mentor's note: Split off from this thread]

No, they didn't. They created six atoms that were in a single quantum state that is a superposition of spin-up and spin-down.

I hope I am not hijacking this thread, but I noticed something. When a system is in a superposition of quantum states, as in the six atoms example, isn't that an expression that, from the probability viewpoint, they do not know what state it is in? What the experiment was saying is that, quantum mechanically, they did not know what state it was in and it could equally likely be in the spin up or the spin down state. Saying it is in a superposition of states is merely a mathematical way of saying we don't know what state it is in and the amplitude of the superposed states just tells us the likelihood of being in one or the other?[/URL]

 

[vbrasic]

I hope I am not hijacking this thread, but I noticed something. When a system is in a superposition of quantum states, as in the six atoms example, isn't that an expression that, from the probability viewpoint, they do not know what state it is in?

Basically. We won't know until we make a measurement, and then, the superposition state will just collapse into any of its constituent states. The superposition state, as you correctly posit, simply tells us what we might see (in your case up/down spins). As you will note, this is indeed a very probabilistic statement.

Saying it is in a superposition of states is merely a mathematical way of saying we don't know what state it is in and the amplitude of the superposed states just tells us the likelihood of being in one or the other?

Well, to be super technical, it tells us the likelihood of being in one of the states IF we measure the system. Otherwise, it just remains in the superposition state. But yes, you are correct, it just gives us a probability.

 

[Strilanc]

Superpositions aren't probability distributions. They don't represent uncertainty.

When a qubit is in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, there's no uncertainty. The von neumann entropy is zero. We know the exact state. There are measurements that you can perform on a qubit in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$ that have uncertain results, but the state itself is not uncertain. It's exactly $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$.

Rational agents can't even disagree on the pure superposition that a system is in (but they can disagree on the mixed state, since those can represent uncertainty). If Alice says the state is $\sqrt{\frac 3 5} |0\rangle + \sqrt{\frac 4 5} |1\rangle$ and Bob says the state is $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, at least one of them just unnecessarily risked losing infinite Bayes points. That's just not the kind of thing that rational agents do.

 

[mike1000]

Superpositions aren't probability distributions. They don't represent uncertainty.

When a qubit is in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, there's no uncertainty. The von neumann entropy is zero. We know the exact state. There are measurements that you can perform on a qubit in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$ that have uncertain results, but the state itself is not uncertain. It's exactly $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$.

Rational agents can't even disagree on the pure superposition that a system is in (but they can disagree on the mixed state, since those can represent uncertainty). If Alice says the state is $\sqrt{\frac 3 5} |0\rangle + \sqrt{\frac 4 5} |1\rangle$ and Bob says the state is $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, at least one of them just unnecessarily risked losing infinite Bayes points. That's just not the kind of thing that rational agents do.

When you make a measurement do you ever actually measure a state which is a superposition of the possible eigenstates?

 

[Strilanc]

When you make a measurement do you ever actually measure a state which is a superposition of the possible eigenstates?

Of course you may. That would be an example of a measurement that has an uncertain outcome. But it's the measurement that's uncertain, not the state. We can be sure about what the state is.

 

[PeterDonis]

We won't know until we make a measurement, and then, the superposition state will just collapse into any of its constituent states.

Only on a collapse interpretation. There are QM interpretations that don't have collapse, such as the MWI.

 

[mike1000]

Of course you may. That would be an example of a measurement that has an uncertain outcome. But it's the measurement that's uncertain, not the state. We can be sure about what the state is.

Well, is the state that you measured a superposition of eigenstates? I am no expert, but if the system can only exist in one of the allowed eigenstates, it seems to me that when you make a measurement you are going to find the system is, indeed, in one of those states. Before hand, when you do not know what state is going to be in, the wave function might be represented by a superposition of eigenstates, each with an amplitude, that is proportional, in some way, to the likelihood of the measurement being in one of the superposed states.

 

[PeterDonis]

if the system can only exist in one of the allowed eigenstates

That is not the case. A system can exist in any state in its Hilbert space. The eigenstates of a particular measurement operator are only a very small subset of all the possible states of the system.

 

[mike1000]

That is not the case. A system can exist in any state in its Hilbert space. The eigenstates of a particular measurement operator are only a very small subset of all the possible states of the system.

When you say in its Hilbert Space, you mean the Hilbert Space of the operator? And would that mean any state which is a linear combination of the eigenstates? Does this mean you can measure the spin of an electron to be both up and down at the same time?

 

[PeterDonis]

When you say in its Hilbert Space, you mean the Hilbert Space of the operator?

No, of the system. The operator is a mapping from the Hilbert space to itself.

would that mean any state which is a linear combination of the eigenstates?

If the measurement operator is complete, then its eigenstates form a basis of the Hilbert space, so any state would be expressible as a linear combination of the eigenstates, yes.

All of this is laid out in QM textbooks; if you haven't studied one, I would try Ballentine.

 

[mike1000]

No, of the system. The operator is a mapping from the Hilbert space to itself.
If the measurement operator is complete, then its eigenstates form a basis of the Hilbert space, so any state would be expressible as a linear combination of the eigenstates, yes.

All of this is laid out in QM textbooks; if you haven't studied one, I would try Ballentine.

Well, does that mean that you can measure the spin of an electron to be both up and down at the same time? I assume that is one of the states in the Hilbert Space of the system.

How do you get the Hilbert Space of the system from the eigenstates of the operator?

 

[PeterDonis]

does that mean that you can measure the spin of an electron to be both up and down at the same time?

Not by measuring spin, no. But see below.

I assume that is one of the states in the Hilbert Space of the system.

Yes; if we have chosen a particular axis for measuring the spin, the "up" and "down" eigenstates of that measurement operator could be written as $\vert \uparrow \rangle$ and $\vert \downarrow \rangle$. The state you describe as "up and down at the same time" (which is actually not a good description of a superposition) would be something like $\frac{1}{\sqrt{2}} \left( \vert \uparrow \rangle + \vert \downarrow \rangle \right)$.

Now, this state is a perfectly valid state, and so there will be some Hermitian operator that has this state as one of its eigenstates. So if we were to measure using that operator (which will not be one of the usual spin operators, of course), we could measure the system to be in this state.

How do you get the Hilbert Space of the system from the eigenstates of the operator?

By forming all possible complex linear combinations of those eigenstates.

 

[PeroK]

Yes; if we have chosen a particular axis for measuring the spin, the "up" and "down" eigenstates of that measurement operator could be written as $\vert \uparrow \rangle$ and $\vert \downarrow \rangle$. The state you describe as "up and down at the same time" (which is actually not a good description of a superposition) would be something like $\frac{1}{\sqrt{2}} \left( \vert \uparrow \rangle + \vert \downarrow \rangle \right)$.

Now, this state is a perfectly valid state, and so there will be some Hermitian operator that has this state as one of its eigenstates.

If the up and down states are in the z-direction, then the operator $S_x$ would have that combination as an eigenstate!

 

[PeterDonis]

If the up and down states are in the z-direction, then the operator $S_x$ would have that combination as an eigenstate!

Yes, good point. So actually, the answer to the question is "yes" even more simply: the state "spin up about the x axis" is in fact a state which is both "up" and "down" about the z axis (if we allow ourselves to use that sloppy language).

Actually, now that I think of it, if we are just considering spin, any linear combination of eigenstates of some spin operator will be an eigenstate of some other spin operator. We would have to consider more general states (for example, states where both spin and position degrees of freedom are present) to find cases where linear combinations of eigenstates of a "simple" operator (like spin or position) are not also eigenstates of a "simple" operator.

 

[mike1000]

Yes, good point. So actually, the answer to the question is "yes" even more simply: the state "spin up about the x axis" is in fact a state which is both "up" and "down" about the z axis (if we allow ourselves to use that sloppy language).

Actually, now that I think of it, if we are just considering spin, any linear combination of eigenstates of some spin operator will be an eigenstate of some other spin operator. We would have to consider more general states (for example, states where both spin and position degrees of freedom are present) to find cases where linear combinations of eigenstates of a "simple" operator (like spin or position) are not also eigenstates of a "simple" operator.

My question is about measurement, what we would measure in an experiment. It seems rather simple, even to someone at my level, that one cannot measure the spin of an electron to be both up and down at the same time, it will be either up or it will be down, not both. I suppose you can manipulate the basis so that up spin or down spin do not coincide with an actual basis direction, but that, it seems to me, is exactly the kind of stuff we should try to avoid. It is that kind of discussion that leads lay people, (like myself) to misunderstand what quantum mechanics is doing, such as the misunderstanding that a system can be in two states at the same time. Some things are simple. Let's not intentionally make them more complex than they already are.

 

[PeterDonis]

My question is about measurement, what we would measure in an experiment.

And if you take a particle in the state I wrote down (in the spin-z basis), and measure its spin about the z axis, you will measure either spin up or spin down. If you take a particle in that state and measure its spin about the x axis, you will always measure spin up--that state is an eigenstate of spin-x with the eigenvalue "up". But given the way that state is written in the spin-z basis, you could also interpret that spin-x measurement as telling you that the particle has z spin "both up and down at the same time". I don't think this would be a very fruitful interpretation, but mathematically how are you going to refute it? The state is what it is, and if you insist on using the spin-z basis it is a superposition, and remains a superposition after you do a spin-x measurement.

It seems rather simple, even to someone at my level, that one cannot measure the spin of an electron to be both up and down at the same time

See above.

I suppose you can manipulate the basis so that up spin or down spin do not coincide with an actual basis direction, but that, it seems to me, is exactly the kind of stuff we should try to avoid

Why? The basis I use will depend on the measurement operator I want to apply, which will depend on the experiment I want to run. Why should I prevent myself from running the experiment I want to run, just because the state of particles coming into the experiment is or is not an eigenstate of the measurement operator?

 

[mike1000]

And if you take a particle in the state I wrote down (in the spin-z basis), and measure its spin about the z axis, you will measure either spin up or spin down. If you take a particle in that state and measure its spin about the x axis, you will always measure spin up--that state is an eigenstate of spin-x with the eigenvalue "up". But given the way that state is written in the spin-z basis, you could also interpret that spin-x measurement as telling you that the particle has z spin "both up and down at the same time". I don't think this would be a very fruitful interpretation, but mathematically how are you going to refute it? The state is what it is, and if you insist on using the spin-z basis it is a superposition, and remains a superposition after you do a spin-x measurement.
See above.
Why? The basis I use will depend on the measurement operator I want to apply, which will depend on the experiment I want to run. Why should I prevent myself from running the experiment I want to run, just because the state of particles coming into the experiment is or is not an eigenstate of the measurement operator?

You are obviously the expert, but I think you over complicate the situation because you want to win a debate. You even admit it is not fruitful. And you are right, it is not fruitful, it is actually harmful. Because you and even I know, that the spin can only be either up or down but not both. I would say this. If the mathematics can be interpreted in a way you describe above, that the electron spin can be both up and down at the same time, then that shows the model is not quite exactly right.

 

[PeterDonis]

Because you and even I know, that the spin can only be either up or down but not both.

No, I don't know that as you state it, because as you state it, it makes it seem like the only possible states of the system are the eigenstates of one particular spin operator. Which is how this whole subthread got started in the first place.

If you really want to be clear, use math. Math is unambiguous. If you absolutely have to have an ordinary language description, you're going to have to make sure to make it as unambiguous and minimal as possible. Something like this, perhaps: "Whenever you measure the spin of a spin-1/2 system, you will get one of two results: up or down." And then stop. Don't say anything about what the state is, or whether it "can be up or down at the same time", or anything like that. Because anything you say, beyond the bare minimum, can and will be misinterpreted. That's why physicists use math when they actually want to make predictions.

If the mathematics can be interpreted in a way you describe above, that the electron spin can be both up and down at the same time, then that shows the model is not quite exactly right

The model makes correct predictions, regardless of how it is interpreted or whether the interpretation is considered fruitful. All that is shown by the possibility of bad interpretations, IMO, is the possibility of bad interpretations. The solution, IMO, is to avoid interpretations altogether. But very few people are willing to do that.

 

[PeterDonis]

I think you over complicate the situation because you want to win a debate.

I would say that you are overcomplicating the situation by trying to overlay an imprecise ordinary language description on top of a precise physical model.

 

[mike1000]

No, I don't know that as you state it, because as you state it, it makes it seem like the only possible states of the system are the eigenstates of one particular spin operator. Which is how this whole subthread got started in the first place.

If you really want to be clear, use math. Math is unambiguous. If you absolutely have to have an ordinary language description, you're going to have to make sure to make it as unambiguous and minimal as possible. Something like this, perhaps: "Whenever you measure the spin of a spin-1/2 system, you will get one of two results: up or down." And then stop. Don't say anything about what the state is, or whether it "can be up or down at the same time", or anything like that. Because anything you say, beyond the bare minimum, can and will be misinterpreted. That's why physicists use math when they actually want to make predictions.
The model makes correct predictions, regardless of how it is interpreted or whether the interpretation is considered fruitful. All that is shown by the possibility of bad interpretations, IMO, is the possibility of bad interpretations. The solution, IMO, is to avoid interpretations altogether. But very few people are willing to do that.

As you have shown, math is not unambiguous. It can be ambiguous too.

I like math very much and I really like to write equations in these posts using the equation editor. As I progress I will write some equations.

 

[PeterDonis]

I have a very hard time learning from you. If I were to let you be my guide, I would never get anywhere. You always find any little thing wrong and you fail to re

It looks like your post got cut off here. In any case, I'm sorry you're frustrated, but I don't think the things I'm pointing out are "little things". You appear to agree that misinterpretation of QM is frequent. So it seems like trying to reduce the chance of misinterpretation would be a good thing. I'm just telling you what I think that requires. Yes, it's pretty stringent--but then again, if it weren't, I wouldn't expect it to fix a problem that is frequent.

As you have shown, math is not unambiguous.

Where did I show that?

 

[mike1000]

It looks like your post got cut off here. In any case, I'm sorry you're frustrated, but I don't think the things I'm pointing out are "little things". You appear to agree that misinterpretation of QM is frequent. So it seems like trying to reduce the chance of misinterpretation would be a good thing. I'm just telling you what I think that requires. Yes, it's pretty stringent--but then again, if it weren't, I wouldn't expect it to fix a problem that is frequent.
Where did I show that?

Ignore the first part that was cut off. I was going to write something and decided not to so you can ignore that part.

 

[bhobba]

I would say that you are overcomplicating the situation by trying to overlay an imprecise ordinary language description on top of a precise physical model.

Exactly.

These things are much better expressed in math.

When Feynman met his hero Dirac - you know what Dirac said - 'I have an equation do you have one to'. Its amusing but contains much truth.

I suggest the following relatively cheap texts:
https://www.amazon.com/dp/0471827223/?tag=pfamazon01-20
https://www.amazon.com/dp/0465075681/?tag=pfamazon01-20
https://www.amazon.com/dp/0465062903/?tag=pfamazon01-20

Thanks
Bill

 

[mike1000]

Exactly.

These things are much better expressed in math.

When Feynman met his hero Dirac - you know what Dirac said - 'I have an equation do you have one to'. Its amusing but contains much truth.

I suggest the following relatively cheap texts:
https://www.amazon.com/dp/0471827223/?tag=pfamazon01-20
https://www.amazon.com/dp/0465075681/?tag=pfamazon01-20
https://www.amazon.com/dp/0465062903/?tag=pfamazon01-20

Thanks
Bill

Yes things are better expressed in math but that is no excuse for not expressing things clearly and simply in english. Here is a paper by Dirac that I found today. What impressed me was the simplicity with which he expressed his thoughts in the first few pages. You have to do both.

http://www.imotiro.org/repositorio/howto/artigoshistoricosordemcronologica/1927%20-%20DIRAC%201927%20First%20steps%20in%20quantum%20field%20theory%20Invention%20of%20the%20second%20quantization%20method.pdf

 

[bhobba]

Yes things are better expressed in math but that is no excuse for not expressing things clearly and simply in english. Here is a paper by Dirac that I found today. What impressed me was the simplicity with which he expressed his thoughts in the first few pages. You have to do both.http://www.imotiro.org/repositorio/howto/artigoshistoricosordemcronologica/1927%20-%20DIRAC%201927%20First%20steps%20in%20quantum%20field%20theory%20Invention%20of%20the%20second%20quantization%20method.pdf

Well I think you have been given clear answers - its just your background makes them not as easy to understand as they should be.

If you read the books I listed you will understand things a lot better.

Please take the time and effort to do it.

Thanks
Bill

 

[mike1000]

Well I think you have been given clear answers - its just your background makes them not as easy to understand as they should be.

If you read the books I listed you will understand things a lot better.

Please take the time and effort to do it.

Thanks
Bill

Please go back and read the first post in this thread. I think I was on the right track.

 

[PeterDonis]

Here is a paper by Dirac that I found today. What impressed me was the simplicity with which he expressed his thoughts in the first few pages.

Yes, agreed. But note that his clear and simple language is describing a precise mathematical model. He makes no statements that are not directly tied to that precise model and its predictions. That is exactly the kind of restraint that I am advocating.

 

[bhobba]

Please go back and read the first post in this thread. I think I was on the right track.

I don't want to interfere with what Peter is doing, but I have to say from a personal view I find explaining superposition very difficult unless you know Linear Algebra. If you do its dead simple - it simply says states form a vector space. But unless you understand Linear Algebra its gibberish.

I have been in many threads discussing it and failed miserably.

Your first post is not on the right track - but I am not the one to explain it because all I would say is the states form a vector space.

As an application see the following on the double slit:
https://arxiv.org/abs/quant-ph/0703126

The double slit is not a demonstration of wave-particle duality which was done away with in the early days of QM. What's really going on from an advanced standpoint is bit involved, but at the beginner level the above paper is much better than the usual wave-particle stuff.

Its simply a demonstration of:
1, The uncertainty principle
2 The principle of superposition. The interference pattern is the result of the superposition of the state behind each slit.

Thanks
Bill

 

[PeterDonis]

Please go back and read the first post in this thread. I think I was on the right track.

I'm afraid I have to agree with bhobba that the first post was not on the right track. Strilanc's responses in posts #3 and #5 look correct to me.

 

[Nugatory]

Because we've come back to the first post in the thread...

I hope I am not hijacking this thread, but I noticed something. When a system is in a superposition of quantum states, as in the six atoms example, isn't that an expression that, from the probability viewpoint, they do not know what state it is in? What the experiment was saying is that, quantum mechanically, they did not know what state it was in and it could equally likely be in the spin up or the spin down state. Saying it is in a superposition of states is merely a mathematical way of saying we don't know what state it is in and the amplitude of the superposed states just tells us the likelihood of being in one or the other?

It is very tempting to think of superposition in this way, but that is not how it works. Consider two different ways of preparing a beam of particles:
1) I start with a bunch of particles in a totally unknown spin state, and I pass them through a horizontally oriented Stern-Gerlach device. This produces two output beams, one of particles in the spin-left state $|L\rangle$ and the other of particles in the spin-right state $|R\rangle$. Now I merge the two beams back together to produce a beam of particles, each of which is either spin-left or spin-right but I don't know which. This state is not a superposition of $|L\rangle$ and $|R\rangle$ (and there is no wave function that describes this state - it's not a vector in a Hilbert space and we need a more general mathematical object called a "density matrix" to represent it) but it is what you are describing.
2) I start with the same bunch of particles in a totally unknown spin state, and I pass them through a vertically oriented Stern-Gerlach device. I keep only the output beam which is deflected upwards, so I have a beam of particles all of which are in the spin-up state $|U\rangle$. This state is a superposition of $|L\rangle$ and $|R\rangle$: $|U\rangle=\frac{\sqrt{2}}{2}(|L\rangle+|R\rangle)$. It is also a superposition, with different coefficients, of $|60\rangle$ and $|240\rangle$ (the states of the particles in the two beams coming out of a Stern-Gerlach device set at 60 degrees to the vertical) and of any other pair of angles I might choose.

If I measure the horizontal spin of the two groups I will get the same result: 50% of my measurements will be spin-left and 50% will be spin-right, so in that case the superposition looks like "left or right, but we don't know which". However, if I measure the spin in any other direction I will get different results. For example, if I measure the first group on the vertical axis, 50% of my measurements will be spin-up and 50% will be spin-down; but if I measure the second group on the vertical axis I will get spin-up every time.

Clearly it won't work to say that the particles prepared in the second way are spin-up before we measure them AND they are either spin-left or spin-right before we measure them but we don't know which AND they are either spin-60 or spin-240 before we measure them but we don't know which. Instead, knowing that they have been prepared in a particular state allows me to predict what I will get if I measure the spin in any given direction. But unless and until I make that measurement, I cannot sensibly claim that the particle has any spin in any direction.

 

[nashed]

My question is about measurement, what we would measure in an experiment. It seems rather simple, even to someone at my level, that one cannot measure the spin of an electron to be both up and down at the same time, it will be either up or it will be down, not both. I suppose you can manipulate the basis so that up spin or down spin do not coincide with an actual basis direction, but that, it seems to me, is exactly the kind of stuff we should try to avoid. It is that kind of discussion that leads lay people, (like myself) to misunderstand what quantum mechanics is doing, such as the misunderstanding that a system can be in two states at the same time. Some things are simple. Let's not intentionally make them more complex than they already are.

I'm answering from my phone so sorry for the lack of formatting, anyway I think you're looking at it from the wrong perspective, in the example of the particle spinning both up and down you are correct to say that if we measure the spin along the up or down direction it will turn up to be either up or down, but if you decide to instead check if it's spinning right or left, it will always be right, that is, instead of saying the electron is spinning both up and down you can say that the electron is spinning right and that's what PeterDonis has been trying to tell you.

For any state you can say that it's doing some things at the same time but you always can find something that it's doing definitely and just say that it's doing that.

 

[Ostrados]

Superposition is a mathematical description of a quantum state (ex: linear combination of eigenstates), saying that a system is in all states simultaneously before measurement is a matter of interpretation (ex: in Bohmian Mechanics particles have definite state from start and there is no superposition).

 

[Karolus]

Superpositions aren't probability distributions. They don't represent uncertainty.

When a qubit is in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, there's no uncertainty. The von neumann entropy is zero. We know the exact state. There are measurements that you can perform on a qubit in the state $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$ that have uncertain results, but the state itself is not uncertain. It's exactly $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$.

Rational agents can't even disagree on the pure superposition that a system is in (but they can disagree on the mixed state, since those can represent uncertainty). If Alice says the state is $\sqrt{\frac 3 5} |0\rangle + \sqrt{\frac 4 5} |1\rangle$ and Bob says the state is $\sqrt{\frac 1 2} |0\rangle + \sqrt{\frac 1 2} |1\rangle$, at least one of them just unnecessarily risked losing infinite Bayes points. That's just not the kind of thing that rational agents do.

So, if I understand correctly, what he says equation (vonNeuman, etc) the qubit is in fact a superposition of states | 0> and | 1>. So we can say that the qubit is simultaneously in the states | 0> and | 1>, at least until we perform a measurement.

 

[vanhees71]

Well, does that mean that you can measure the spin of an electron to be both up and down at the same time? I assume that is one of the states in the Hilbert Space of the system.

How do you get the Hilbert Space of the system from the eigenstates of the operator?

Again and again and again, please don't use this nonsense talk about a system being in more than one state or an observable taking more than one value at the same time. That's NOT implied by anything in quantum theory, but it's claimed by bad popular-science texts, and there are more bad than good popular-science writings around. The problem is that to write a good popular-science article is among the most difficult tasks for a scientist; in some sense it's more difficult than to write a scientific article, because popular-science writing forbids one to use the only language that is really adequate to talk about quantum theory, i.e., math!

Quantum theory tells you that a system is always in one state, which can be proper or mixed. Even in a completely determined state, i.e., if the system is prepared such that a complete set of compatible observables takes determined values, other observables (that are not compatible with all observables of the complete set) do not necessarily take determined values, but than it's wrong to say that they take more than one value at once. In this case they simply don't take any determined value. The state, implies the probabilities to measure any possible value of that observable and nothing else. To explain this in a really understandable way, I'd have to use the mathematical formalism of quantum theory.

 

[Strilanc]

So, if I understand correctly, what he says equation (vonNeuman, etc) the qubit is in fact a superposition of states | 0> and | 1>. So we can say that the qubit is simultaneously in the states | 0> and | 1>, at least until we perform a measurement.

No, it's in the state $\frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$ (for example).

Sometimes people describe the state $\frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$ as "simultaneously in both 0 and 1", but this is an just an oversimplified and misleading translation from the math into English. It's like saying that a diagonal line is "simultaneously both a horizontal line and a vertical line". That's a terrible description of what it means to be diagonal.